$C^\infty$ Periodic Pole-free Rational Interpolation

Question

let $\quad-1=x_0 < x_1 <\ ...\ < x_n<1\quad$ be a set of abscissas
and $\quad(y_0, y_1,\ ...\,y_n)\quad$ a sequence of the corresponding ordinates.

Question:

what can be said about the existence and calculation of a pole-free rational function $R(x)$,
with the following properties?:

$\quad R(x_i) = y_i,\quad\quad\quad 0 \le i\le n$

$\quad \frac{d^îR}{dx^î}(1) = \frac{d^iR}{dx^i}(-1),\quad \forall i\in\mathbb{N}_0, \left(\frac{d^0R}{dx^0}(x) := R(x)\right)$

Background of my question is Barycentric Rational Interpolation is an infinitely often differentable analytic alternative to Splines and NURBS and I wonder, if rational interpolation could also provide infinitely often differentiable analytic closed curves through a set of points e.g. in the plane.

Answer 1

No such function exists, except possibly when $R(x)$ is a constant. Your condition on the derivatives implies periodicity (the periodic extensions of $R(x)$ from $[-1,1]$ to the whole real line is analytic and agrees with $R(x)$ on $[-1,1]$ and must be equal to $R(x)$ by uniqueness of analytic continuation). No non-constant rational function is periodic (if $R(x)$ is rational, you can find a positive integer $n$ such that $x^n R(1/x)$ is analytic at $x=0$, and that is not true for any periodic analytic function other than a constant).